91Ó°ÊÓ

Let's denote the number of deaths per week as X and the number of weeks when X deaths occurred as f. To compute the mean number of deaths per week, we will use the following formula: Mean, \(\mu = \frac{\sum{X \cdot f}}{\sum{f}}\). Using the provided data: \(\mu = \frac{(0 \cdot 109) + (1 \cdot 65) + (2 \cdot 22) + (3 \cdot 3) + (4 \cdot 1)}{109 + 65 + 22 + 3 + 1} = \frac{0 + 65 + 44 + 9 + 4}{200} = \frac{122}{200}\) So, \(\mu = 0.61\).

Since we have calculated the mean number of deaths per week (\(\mu = 0.61\)), we can now determine the Poisson distribution. The formula for the Poisson distribution is: \(P(X=k) = \frac{e^{-\mu} \cdot \mu^{k}}{k!}\) Where k is the number of deaths per week, e is the base of the natural logarithm (approximately 2.71828), and \(k!\) is the factorial of k.

Now, we will use the determined Poisson distribution to calculate the predicted frequencies for the given data. The predicted frequency for each value of k can be found by using the formula: Predicted frequency = \(P(X=k) \cdot \sum{f}\) Considering that there are 200 weeks in total, we can calculate the predicted frequencies for each value of k from 0 to 4: \(P(X=0) = \frac{e^{-0.61} \cdot 0.61^{0}}{0!} = e^{-0.61}\) Predicted frequency of 0 deaths: \( e^{-0.61} \cdot 200\) \(P(X=1) = \frac{e^{-0.61} \cdot 0.61^{1}}{1!}\) Predicted frequency of 1 death: \(\frac{e^{-0.61} \cdot 0.61}{1} \cdot 200\) \(P(X=2) = \frac{e^{-0.61} \cdot 0.61^{2}}{2!}\) Predicted frequency of 2 deaths: \(\frac{e^{-0.61} \cdot 0.61^{2}}{2} \cdot 200\) \(P(X=3) = \frac{e^{-0.61} \cdot 0.61^{3}}{3!}\) Predicted frequency of 3 deaths: \(\frac{e^{-0.61} \cdot 0.61^{3}}{6} \cdot 200\) \(P(X=4) = \frac{e^{-0.61} \cdot 0.61^{4}}{4!}\) Predicted frequency of 4 deaths: \(\frac{e^{-0.61} \cdot 0.61^{4}}{24} \cdot 200\)

We'll now compare the observed frequencies (given in the table) with the predicted frequencies calculated in step 3: Observed frequencies: 109 (0 deaths), 65 (1 death), 22 (2 deaths), 3 (3 deaths), 1 (4 deaths). Predicted frequencies (rounded to the nearest whole number): - 0 deaths: \( e^{-0.61} \cdot 200 \approx 108\) - 1 death: \(\frac{e^{-0.61} \cdot 0.61}{1} \cdot 200 \approx 66\) - 2 deaths: \(\frac{e^{-0.61} \cdot 0.61^{2}}{2} \cdot 200 \approx 20\) - 3 deaths: \(\frac{e^{-0.61} \cdot 0.61^{3}}{6} \cdot 200 \approx 4\) - 4 deaths: \(\frac{e^{-0.61} \cdot 0.61^{4}}{24} \cdot 200 \approx 1\) Comparing these values, we can see that the predicted frequencies for deaths per week resulting from automobile accidents are very close to the observed frequencies. This indicates that the Poisson distribution is a good fit for this dataset.